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3 years ago

10

Which of the filling is a result of the transfer of energy?

1 answer:

bija089 [108]3 years ago

6 0

Explanation:

the speed of sound depends on the particles of the medium of the sound wave is moving through because the particles in the solid and tightly packed together South Wales weekly transfers energy from particle to particle in a solid

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How do I find distance traveled

Marianna [84]

Answer:

To solve for distance use the formula for distance d = st, or distance equals speed times time. Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t.

Explanation:

5 0

3 years ago

Suppose a lion running at a velocity of 10 m/s east slows

iris [78.8K]

Answer:

$a=0.34\ m/s^2$

Explanation:

Given that,

Initial velocity, u = 10 m/s

Final velocity, v = 0 (stops)

Time, t = 30 s

We need to find the acceleration of the lion.

We know that,

Acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\a=\dfrac{10-0}{30}\\\\a=0.34\ m/s^2$

So, the acceleration of the lion is $0.34\ m/s^2$ .

5 0

3 years ago

Which distance–time graph most closely represents an object moving with uniform motion? and for the second one, Which speed–time

Georgia [21]

Answer:

1. B

2. D

Explanation:

7 0

4 years ago

The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his

Alex17521 [72]

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

$F_x=kx$

Substitute the values

$6500=1300x$

$x=\frac{6500}{1300}=5$ m

Work done due to friction force

$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

Initial kinetic energy, Ki=0

Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

Substitute the values

$0+0+16250-250=30v^2+1470+0$

$16000-1470=30v^2$

$14530=30v^2$

$v^2=\frac{14530}{30}$

$v=\sqrt{\frac{14530}{30}}$

$v=22m/s$

5 0

3 years ago

Please help! Will mark best answer!!

MA_775_DIABLO [31]

The answer is (A) hope it helps

7 0

3 years ago

Read 2 more answers